/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 A system executes a power cycle ... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 1

A system executes a power cycle while receiving \(1000 \mathrm{~kJ}\) by heat transfer at a temperature of \(500 \mathrm{~K}\) and discharging energy by heat transfer at a temperature of \(300 \mathrm{~K}\). There are no other heat transfers. Applying Eq. 6.2, determine \(\sigma_{\text {cycle }}\) if the thermal efficiency is (a) \(60 \%\), (b) \(40 \%\), (c) \(20 \%\). Identify the cases (if any) that are internally reversible or impossible.

Short Answer

Expert verified
(a) Impossible, \(\sigma_{\text{cycle}} = -\frac{2}{3}\). (b) Internally reversible, \(\sigma_{\text{cycle}} = 0\). (c) Irreversible, \(\sigma_{\text{cycle}} = \frac{2}{3}\).

Step by step solution

01

Identify Given Information

The system executes a power cycle with the following details: - Heat input, \(Q_H = 1000 \mathrm{~kJ}\) at \(T_H = 500 \mathrm{~K}\).- Heat rejection at \(T_C = 300 \mathrm{~K}\).- No other heat transfers.We need to determine \(\sigma_{\text {cycle }}\) using the given thermal efficiencies of 60%, 40%, and 20% and then identify whether each case is internally reversible or impossible.
02

Calculate Useful Work Output

Thermal efficiency \(\eta\) is given by \[ \eta = \frac{W_{net}}{Q_H} \]Rearrange to find \(W_{net}\): \[ W_{net} = \eta \cdot Q_H \]
03

Determine the Heat Rejected, \(Q_L\)

Using the first law of thermodynamics, the heat rejected \(Q_L\) can be found using the equation:\[ Q_H = W_{net} + Q_L \]Rearranging, we get:\[ Q_L = Q_H - W_{net} \]
04

Calculate Entropy Generation, \(\sigma_{\text{cycle}}\)

Using Eq. 6.2, the entropy generation \(\sigma_{\text{cycle}}\) is given by: \[ \sigma_{\text{cycle}} = \frac{Q_L}{T_C} - \frac{Q_H}{T_H} \]
05

Case (a) - Calculate for 60% Efficiency

Given \(\eta = 0.60\):\[ W_{net} = 0.60 \cdot 1000 \mathrm{~kJ} = 600 \mathrm{~kJ} \]\[ Q_L = 1000 \mathrm{~kJ} - 600 \mathrm{~kJ} = 400 \mathrm{~kJ} \]Now, calculate entropy generation:\[ \sigma_{\text{cycle}} = \frac{400}{300} - \frac{1000}{500} = \frac{4}{3} - 2 = -\frac{2}{3} \]Since \(\sigma_{\text{cycle}} < 0\), this process is impossible.
06

Case (b) - Calculate for 40% Efficiency

Given \(\eta = 0.40\):\[ W_{net} = 0.40 \cdot 1000 \mathrm{~kJ} = 400 \mathrm{~kJ} \]\[ Q_L = 1000 \mathrm{~kJ} - 400 \mathrm{~kJ} = 600 \mathrm{~kJ} \]Now, calculate entropy generation:\[ \sigma_{\text{cycle}} = \frac{600}{300} - \frac{1000}{500} = 2 - 2 = 0 \]Since \(\sigma_{\text{cycle}} = 0\), this process is internally reversible.
07

Case (c) - Calculate for 20% Efficiency

Given \(\eta = 0.20\):\[ W_{net} = 0.20 \cdot 1000 \mathrm{~kJ} = 200 \mathrm{~kJ} \]\[ Q_L = 1000 \mathrm{~kJ} - 200 \mathrm{~kJ} = 800 \mathrm{~kJ} \]Now, calculate entropy generation:\[ \sigma_{\text{cycle}} = \frac{800}{300} - \frac{1000}{500} = \frac{8}{3} - 2 = \frac{2}{3} \]Since \(\sigma_{\text{cycle}} > 0\), this process is irreversible.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

entropy generation
Entropy generation \( \sigma_{\text {cycle}} \) is a key concept in thermodynamics, helping us understand the irreversibility of processes. In simple terms, entropy generation measures how much disorder or randomness increases due to a process. For a reversible process, the entropy generation is zero, meaning no irreversibility.
As entropy generation increases, the process becomes more irreversible. It's important to remember that a negative entropy generation is impossible as it would violate the second law of thermodynamics.
When calculating \( \sigma_{\text {cycle}} \) for a power cycle, we need the heat rejected ( \( \ Q_L \ \) ) and the heat input ( \( \ Q_H \ \) ) along with the respective temperatures at which these heat transfers occur:
\( \sigma_{\text{cycle}} = \frac{Q_L}{T_C} - \frac{Q_H}{T_H} \ \textrm{ .} \)
From the exercise, we can see the calculation for each thermal efficiency case:
  • For 60% efficiency, entropy generation is negative, making the process impossible.
  • For 40% efficiency, entropy generation is zero, indicating an internally reversible process.
  • For 20% efficiency, entropy generation is positive, showing an irreversible process.
thermal efficiency
Thermal efficiency \( \eta \ \) is a measure of how well a power cycle converts heat into work. It is a ratio defined by:
\( \eta = \frac{W_{net}}{Q_H} \)
\

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Most popular questions from this chapter

Ammonia enters a valve as a saturated liquid at 7 bar with a mass flow rate of \(0.06 \mathrm{~kg} / \mathrm{min}\) and is steadily throttled to a pressure of 1 bar. Determine the rate of entropy production in \(\mathrm{kW} / \mathrm{K}\). If the valve were replaced by a power-recovery turbine operating at steady state, determine the maximum theoretical power that could be developed, in \(\mathrm{kW}\). In each case, ignore heat transfer with the surroundings and changes in kinetic and potential energy. Would you recommend using such a turbine?

A reversible refrigeration cycle \(\mathrm{R}\) and an irreversible refrigeration cycle I operate between the same two reservoirs and each removes \(Q_{\mathrm{C}}\) from the cold reservoir. The net work input required by \(\mathrm{R}\) is \(W_{\mathrm{R}}\), while the net work input for \(\mathrm{I}\) is \(W_{\mathrm{I}}\). The reversible cycle discharges \(Q_{\mathrm{H}}\) to the hot reservoir, while the irreversible cycle discharges \(Q_{\mathrm{H}}^{\prime}\). Show that \(W_{1}>W_{\mathrm{R}}\) and \(Q_{\mathrm{H}}^{\prime}>Q_{\mathrm{H}}\).

Carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) enters a nozzle operating at steady state at 28 bar, \(267^{\circ} \mathrm{C}\), and \(50 \mathrm{~m} / \mathrm{s}\). At the nozzle exit, the conditions are \(1.2\) bar, \(67^{\circ} \mathrm{C}, 580 \mathrm{~m} / \mathrm{s}\), respectively. (a) For a control volume enclosing the nozzle only, determine the heat transfer, in \(\mathrm{kJ}\), and the change in specific entropy, in \(\mathrm{kJ} / \mathrm{K}\), each per \(\mathrm{kg}\) of carbon dioxide flowing through the nozzle. What additional information would be required to evaluate the rate of entropy production? (b) Evaluate the rate of entropy production, in \(\mathrm{kJ} / \mathrm{K}\) per \(\mathrm{kg}\) of carbon dioxide flowing, for an enlarged control volume enclosing the nozzle and a portion of its immediate surroundings so that the heat transfer occurs at the ambient temperature, \(25^{\circ} \mathrm{C}\).

A gas flows through a one-inlet, one-exit control volume operating at steady state. Heat transfer at the rate \(\dot{Q}_{\mathrm{cv}}\) takes place only at a location on the boundary where the temperature is \(T_{\mathrm{b}}\). For each of the following cases, determine whether the specific entropy of the gas at the exit is greater than, equal to, or less than the specific entropy of the gas at the inlet: (a) no internal irreversibilities, \(\dot{Q}_{\mathrm{cv}}=0\). (b) no internal irreversibilities, \(\dot{Q}_{\mathrm{cv}}<0\). (c) no internal irreversibilities, \(\dot{Q}_{\mathrm{cv}}>0\). (d) internal irreversibilities, \(\dot{Q}_{\mathrm{cv}}<0\). (e) internal irreversibilities, \(\dot{Q}_{\mathrm{cv}} \geq 0\)

Nitrogen \(\left(\mathrm{N}_{2}\right)\) initially occupying \(0.5 \mathrm{~m}^{3}\) at \(1.0\) bar, \(20^{\circ} \mathrm{C}\) undergoes an internally reversible compression during which \(p V^{1.30}=\) constant to a final state where the temperature is \(200^{\circ} \mathrm{C}\). Determine assuming the ideal gas model (a) the pressure at the final state, in bar. (b) the work and heat transfer, each in \(\mathrm{kJ}\). (c) the entropy change, in \(\mathrm{kJ} / \mathrm{K}\).
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